Prove that ∫π/40(√tanx+√cotx)dx=√2⋅π2

Let LHS=∫π/40(√tanx+√cotx)dx=∫π/40(√tanx⎛⎜⎜⎜⎜⎜⎝√sinx√(cosx)+√cosx√sinx⎞⎟⎟⎟⎟⎟⎠dx=∫π/40(√tanxsinx+cosx√sinx.cosxThus√2∫π/40(√tanxsinx+cosx√2sinx.cosxdx=√2∫∫π/40(√tanxsinx+cosx√1−(sinx−cosx)2Letsinx−cosx=z⇒(cosx+sinx)dx=dzAlso Ifx=0,z=𕒵andx=π4,z=1√2𕒵√2=0Therefore, LHS=√2∫0𕒵dz√1−z2=√2[sin𕒵z]0𕒵=√2[sin󔼒−sin𕒵(𕒵)]=√2[0−(−π2)]=√2.π2= RHS.Hence proved.